The scores on a statewide language exam were normally distributed with $\mu = 69.72$ and $\sigma = 4$. Stephanie earned a n $81$ on the exam. Stephanie's exam grade was higher than what fraction of test-takers? Use the cumulative z-table provided below. z.00.01.02.03.04.05.06.07.08.09 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974 2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981 2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986
Explanation: A cumulative z-table shows the probability that a standard normal variable will be less than a certain value (z) In order to use the z-table, we first need to determine the z-score of Stephanie's exam grade. Recall that we can calculate her z-score by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}} = \dfrac{81 - {69.72}}{{4}} = 2.82} $ Look up $2.82$ on the z-table. This value, $0.9976$ , represents the portion of the population that scored lower than $81$ on the exam. Stephanie scored higher than $99.76\%$ of the test-takers on the language exam.